What is the integral from 0 to pi/2 of cos^4(t) ?

The integral from 0 to pi/2 of cos^4(t) is (3pi)/(16)

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integral from 0 to pi/2 of cos^4(t)

\int_{0}^{\frac{π}{2}} cos^{4} (t) d t

\frac{3 π}{16}

Decimal Notation

0.58904 \dots

Solution steps

\int_{0}^{\frac{π}{2}} cos^{4} (t) d t

Simplify

cos^{4} (t) : cos^{3} (t) cos (t)

= \int_{0}^{\frac{π}{2}} cos^{3} (t) cos (t) d t

Apply Integration By Parts

: [cos^{3} (t) sin (t) - \int - 3 cos^{2} (t) sin^{2} (t) d t]_{0}^{\frac{π}{2}}

= [cos^{3} (t) sin (t) - \int - 3 cos^{2} (t) sin^{2} (t) d t]_{0}^{\frac{π}{2}}

\int - 3 cos^{2} (t) sin^{2} (t) d t = - \frac{3}{8} (t - \frac{1}{4} sin (4 t))

= [cos^{3} (t) sin (t) - (- \frac{3}{8} (t - \frac{1}{4} sin (4 t)))]_{0}^{\frac{π}{2}}

Simplify

[cos^{3} (t) sin (t) - (- \frac{3}{8} (t - \frac{1}{4} sin (4 t)))]_{0}^{\frac{π}{2}} : [\frac{1}{32} (32 cos^{3} (t) sin (t) + 3 (4 t - sin (4 t)))]_{0}^{\frac{π}{2}}

= [\frac{1}{32} (32 cos^{3} (t) sin (t) + 3 (4 t - sin (4 t)))]_{0}^{\frac{π}{2}}

Compute the boundaries:

\frac{3 π}{16}

= \frac{3 π}{16}

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What is the integral from 0 to pi/2 of cos^4(t) ?
The integral from 0 to pi/2 of cos^4(t) is (3pi)/(16)